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Difference between revisions of "2004 AIME I Problems/Problem 14"

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== Problem ==
 
== Problem ==
−
A unicorn is tethered by a 20-foot silver rope to the base of a magician's cylindrical tower whose radius is 8 feet. The rope is attached to the tower at ground level and to the unicorn at a height of 4 feet. The unicorn has pulled the rope taut, the end of the rope is 4 feet from the nearest point on the tower, and the length of the rope that is touching the tower is <math> \frac{a-\sqrt{b}}c </math> feet, where <math> a, b, </math> and <math> c </math> are positive integers, and <math> c </math> is prime. Find <math> a+b+c. </math>
+
A unicorn is tethered by a <math>20</math>-foot silver rope to the base of a magician's [[cylinder|cylindrical]] tower whose [[radius]] is <math>8</math> feet. The rope is attached to the tower at ground level and to the unicorn at a height of <math>4</math> feet. The unicorn has pulled the rope taut, the end of the rope is <math>4</math> feet from the nearest point on the tower, and the length of the rope that is touching the tower is <math> \frac{a-\sqrt{b}}c </math> feet, where <math> a, b, </math> and <math> c </math> are [[positive]] [[integer]]s, and <math> c </math> is prime. Find <math> a+b+c. </math>
  
 
== Solution ==
 
== Solution ==
−
Looking from an overhead view, call the center of the circle O, the tether point to the unicorn A, and the last point where the rope touches the tower B.  Triangle OAB is a right triangle because OB is a radius and BA is a tangent line of point B.  We use the Pythagorean theorem to find the horizontal component of AB=(80)^(1/2). Now looking at a side view and "unrolling" the cylinder to be a flat surface, call the bottom tether of the rope C, the point on the ground below A D, and the point directly above B and 4 feet off the ground E.  Triangles CDA and AEB are similar right triangles.  By the Pythagorean theorem CD=8*6^(1/2).  Let the length of AB=x. CA/CD=AB/AE: (5/2)*6^(1/2)=x/80^(1/2).  20-x=(60-750^(1/2))/3, Therefore a=60, b=750, c=3, a+b+c=813.
+
<center>
 +
<asy>
 +
  /* Settings */
 +
import three; defaultpen(fontsize(10)+linewidth(0.62));
 +
currentprojection = perspective(-2,-50,15); size(200);
 +
  /* Variables */
 +
real x = 20 - ((750)^.5)/3, CE = 8*(6^.5) - 4*(5^.5), CD = 8*(6^.5), h = 4*CE/CD;
 +
pair Cxy = 8*expi((3*pi)/2-CE/8);
 +
triple Oxy = (0,0,0), A=(4*5^.5,-8,4), B=(0,-8,h), C=(Cxy.x,Cxy.y,0), D=(A.x,A.y,0), E=(B.x,B.y,0), O=(O.x,O.y,h);
 +
pair L = 8*expi(pi+0.05), R = 8*expi(-0.22); /* left and right cylinder lines, numbers from trial/error */
 +
  /* Drawing */
 +
draw(B--A--D--E--B--C);
 +
draw(circle(Oxy,8));
 +
draw(circle(O,8));
 +
draw((L.x,L.y,0)--(L.x,L.y,h)); draw((R.x,R.y,0)--(R.x,R.y,h));
 +
draw(O--B--(A.x,A.y,h)--cycle,dashed);
 +
  /* Labeling */
 +
label("\(A\)",A,NE);  dot(A);
 +
label("\(B\)",B,NW);  dot(B);
 +
label("\(C\)",C,W);  dot(C);
 +
label("\(D\)",D,E);  dot(D);
 +
label("\(E\)",E,S);  dot(E);
 +
label("\(O\)",O,NW);  dot(O);
 +
 
 +
pair O1 = (25,4), A1=O1+(4*sqrt(5),-8), B1=O1+(0,-8);
 +
draw(circle(O1,8));
 +
draw(O1--A1--B1--O1);
 +
label("\(A\)",A1,SE);label("\(B\)",B1,SW);label("\(O\)",O1,N);
 +
label("$8$",O1--B1,W);label("$4$",(5*A1+O1)/6,0.8*(1,1));
 +
label("$8$",O1--A1,0.8*(-0.5,2));label("$4\sqrt{5}$",B1/2+A1/2,(0,-1));
 +
</asy></center>
 +
Looking from an overhead view, call the [[center]] of the [[circle]] <math>O</math>, the tether point to the unicorn <math>A</math> and the last point where the rope touches the tower <math>B</math>.  <math>\triangle OAB</math> is a [[right triangle]] because <math>OB</math> is a radius and <math>BA</math> is a [[tangent line]] at point <math>B</math>.  We use the [[Pythagorean Theorem]] to find the horizontal component of <math>AB</math> has length <math>4\sqrt{5}</math>. 
 +
 
 +
<center><asy>
 +
defaultpen(fontsize(10)+linewidth(0.62));
 +
pair A=(-4*sqrt(5),4), B=(0,4*(8*sqrt(6)-4*sqrt(5))/(8*sqrt(6))), C=(8*sqrt(6)-4*sqrt(5),0), D=(-4*sqrt(5),0), E=(0,0);
 +
draw(A--C--D--A);draw(B--E);
 +
label("\(A\)",A,(-1,1));label("\(B\)",B,(1,1));label("\(C\)",C,(1,0));label("\(D\)",D,(-1,-1));label("\(E\)",E,(0,-1));
 +
label("$4\sqrt{5}$",D/2+E/2,(0,-1));label("$8\sqrt{6}-4\sqrt{5}$",C/2+E/2,(0,-1));
 +
label("$4$",D/2+A/2,(-1,0));label("$x$",C/2+B/2,(1,0.5));label("$20-x$",0.7*A+0.3*B,(1,0.5));
 +
dot(A^^B^^C^^D^^E);
 +
</asy></center>
 +
 
 +
Now look at a side view and "unroll" the cylinder to be a flat surface. Let <math>C</math> be the bottom tether of the rope, let <math>D</math> be the point on the ground below <math>A</math>, and let <math>E</math> be the point directly below <math>B</math>.  [[Triangle]]s <math>\triangle CDA</math> and <math>\triangle CEB</math> are [[similar]] [[right triangle]]s.  By the Pythagorean Theorem <math>CD=8\cdot\sqrt{6}</math>.   
 +
 
 +
Let <math>x</math> be the length of <math>CB</math>. <cmath>\frac{CA}{CD}=\frac{CB}{CE}\implies \frac{20}{8\sqrt{6}}=\frac{x}{8\sqrt{6}-4\sqrt{5}}\implies x=\frac{60-\sqrt{750}}{3}</cmath>
 +
 
 +
Therefore <math>a=60, b=750, c=3, a+b+c=\boxed{813}</math>.
  
 
== See also ==
 
== See also ==
−
* [[2004 AIME I Problems/Problem 13| Previous problem]]
+
{{AIME box|year=2004|n=I|num-b=13|num-a=15}}
−
 
 
−
* [[2004 AIME I Problems/Problem 15| Next problem]]
 
  
−
* [[2004 AIME I Problems]]
+
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Revision as of 14:51, 16 April 2021

Problem

A unicorn is tethered by a $20$-foot silver rope to the base of a magician's cylindrical tower whose radius is $8$ feet. The rope is attached to the tower at ground level and to the unicorn at a height of $4$ feet. The unicorn has pulled the rope taut, the end of the rope is $4$ feet from the nearest point on the tower, and the length of the rope that is touching the tower is $\frac{a-\sqrt{b}}c$ feet, where $a, b,$ and $c$ are positive integers, and $c$ is prime. Find $a+b+c.$

Solution

[asy] /* Settings */ import three; defaultpen(fontsize(10)+linewidth(0.62)); currentprojection = perspective(-2,-50,15); size(200); /* Variables */ real x = 20 - ((750)^.5)/3, CE = 8*(6^.5) - 4*(5^.5), CD = 8*(6^.5), h = 4*CE/CD; pair Cxy = 8*expi((3*pi)/2-CE/8); triple Oxy = (0,0,0), A=(4*5^.5,-8,4), B=(0,-8,h), C=(Cxy.x,Cxy.y,0), D=(A.x,A.y,0), E=(B.x,B.y,0), O=(O.x,O.y,h); pair L = 8*expi(pi+0.05), R = 8*expi(-0.22); /* left and right cylinder lines, numbers from trial/error */ /* Drawing */ draw(B--A--D--E--B--C); draw(circle(Oxy,8)); draw(circle(O,8)); draw((L.x,L.y,0)--(L.x,L.y,h)); draw((R.x,R.y,0)--(R.x,R.y,h)); draw(O--B--(A.x,A.y,h)--cycle,dashed); /* Labeling */ label("\(A\)",A,NE); dot(A); label("\(B\)",B,NW); dot(B); label("\(C\)",C,W); dot(C); label("\(D\)",D,E); dot(D); label("\(E\)",E,S); dot(E); label("\(O\)",O,NW); dot(O); pair O1 = (25,4), A1=O1+(4*sqrt(5),-8), B1=O1+(0,-8); draw(circle(O1,8)); draw(O1--A1--B1--O1); label("\(A\)",A1,SE);label("\(B\)",B1,SW);label("\(O\)",O1,N); label("$8$",O1--B1,W);label("$4$",(5*A1+O1)/6,0.8*(1,1)); label("$8$",O1--A1,0.8*(-0.5,2));label("$4\sqrt{5}$",B1/2+A1/2,(0,-1)); [/asy]

Looking from an overhead view, call the center of the circle $O$, the tether point to the unicorn $A$ and the last point where the rope touches the tower $B$. $\triangle OAB$ is a right triangle because $OB$ is a radius and $BA$ is a tangent line at point $B$. We use the Pythagorean Theorem to find the horizontal component of $AB$ has length $4\sqrt{5}$.

[asy] defaultpen(fontsize(10)+linewidth(0.62)); pair A=(-4*sqrt(5),4), B=(0,4*(8*sqrt(6)-4*sqrt(5))/(8*sqrt(6))), C=(8*sqrt(6)-4*sqrt(5),0), D=(-4*sqrt(5),0), E=(0,0); draw(A--C--D--A);draw(B--E); label("\(A\)",A,(-1,1));label("\(B\)",B,(1,1));label("\(C\)",C,(1,0));label("\(D\)",D,(-1,-1));label("\(E\)",E,(0,-1)); label("$4\sqrt{5}$",D/2+E/2,(0,-1));label("$8\sqrt{6}-4\sqrt{5}$",C/2+E/2,(0,-1)); label("$4$",D/2+A/2,(-1,0));label("$x$",C/2+B/2,(1,0.5));label("$20-x$",0.7*A+0.3*B,(1,0.5)); dot(A^^B^^C^^D^^E); [/asy]

Now look at a side view and "unroll" the cylinder to be a flat surface. Let $C$ be the bottom tether of the rope, let $D$ be the point on the ground below $A$, and let $E$ be the point directly below $B$. Triangles $\triangle CDA$ and $\triangle CEB$ are similar right triangles. By the Pythagorean Theorem $CD=8\cdot\sqrt{6}$.

Let $x$ be the length of $CB$. \[\frac{CA}{CD}=\frac{CB}{CE}\implies \frac{20}{8\sqrt{6}}=\frac{x}{8\sqrt{6}-4\sqrt{5}}\implies x=\frac{60-\sqrt{750}}{3}\]

Therefore $a=60, b=750, c=3, a+b+c=\boxed{813}$.

See also

2004 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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All AIME Problems and Solutions

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